1

Random variables and probability distributions

2

Random variables are a key concept for statistical inference

We know that in order to use results from a sample to

infer about a larger population, we must avoid bias and

chose a sample at random. If our sample is random,

then the sample mean, the sample proportion, and other

statistics that we calculate from the sample are called:

random variables. The behavior of such variables is an

essential step on the way to performing statistical

inference.

3

Random variablesA random variable X associates a numerical value with each elementary outcome of an experiment

ExampleLet X be the number of boys born in a family of 2 children. List the possible values of X

X=1 means that there is one boy in the family

2 children in the family X – number of boys

BB 2

BG 1

GB 1

GG 0

4

Probability distribution of discrete variables

In order to understand the pattern of behavior of a random variable, we need to know its possible values and how likely they are to occur.The probability distribution of a discrete random variable X is a list of the distinct numerical values of X, along with their associated probabilities.

X x1 x2 x3 …… xk

Probability p1 p2 p3 …… pk

0≤pi≤1 i=1,…,k

p1+p2+…+pk=1

5

Example

If X represents the number of boys in a family of 2 children, find the probability distribution of X.

The probability of BB is:

p(first is a boy and second is a boy) = p(first is a boy)× p(second is a boy)=0.5×0.5=0.25

P(BG)= P(GB)= P(GG)=

Family with 2 children X – number of boys Probability

BB 2

BG 1

GB 1

GG 0

Since the 2 events are independent

0.25 0.25 0.25

Family with 2 children X – number of boys Probability

BB 2 0.25

BG 1 0.25

GB 1 0.25

GG 0 0.25

6

Example - continued

Now we can specify the probability of each value of X

Family with 2 children X – number of boys Probability

BB 2 0.25

BG 1 0.25

GB 1 0.25

GG 0 0.25

Value of X Probability

0 0.25

1 0.25+0.25=0.5

2 0.25

7

Notations for random variables

We denote the random variable by uppercase letters (e.g., X,V,Z)

We denote the values that the random variable gets by lowercase letters (e.g., xi, vi, zi).

The probability that a random variable X get a certain value x is denoted by: p(X=xi), or, in short, p(xi).

For example

Value of X P(xi)

0 0.25

1 0.5

2 0.25

P(X=0)=0.25

P(X=1)=0.5

P(X=2)=0.25

Remember that the sum of the probabilities must equal 1!!!

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Example

A probability distribution is given in the accompanying table with the additional information that the even values of X are equally likely. Determine the missing entries of the table.

Value of X P(x)

1 0.2

2

3 0.2

4

5 0.3

6

Answer:

The sum of the probabilities in the table:

0.2+0.2+0.3=0.7

The remaining 0.3 probability is equally divided between the values 2,4,and 6

0.1

0.1

0.1

9

ExampleHere is the probability distribution p(x) for grade X of a randomly chosen student from a certain class. X=0 represents an F, X=4 represents an A.

X p(x)(F) 0 0.08(D) 1 0.06(C) 2 0.24(B) 3 0.36(A) 4 0.26

1. What is the probability of getting at least a B?

2. What is the probability of getting higher than a D?

3. What is the probability of getting less than a C?

4. What is the probability of getting no better than a B?

1. .62.26.364)p(X3)p(X3)p(X

.86.26.3624.4)p(X3)p(X2)p(X1)p(X

.14.08.060)p(X1)p(X2)p(X

.74.36.24.06.083)p(X2)p(X1)p(X0)p(X3)p(X

2.

3.

4.

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Probability histogram

A probability histogram serves as a display of a probability distribution

ExampleNumber of boys is a family of 2 children

X P(x)

0 0.25

1 0.5

2 0.25

11

Example - rolling two dice:

X=sum of two dice

X=2,3,4,5,6,7,8,9,10,11,12

Outcome of 2 dice X Probability(1,1) 2

3

4

5

6

7

8

9

10

11

12

Outcome of 2 dice X Probability(1,1) 2

(1,2) (2,1) 3

4

5

6

7

8

9

10

11

12

Outcome of 2 dice X Probability(1,1) 2

(1,2) (2,1) 3

(1,3) (2,2) (3,1) 4

5

6

7

8

9

10

11

12

Outcome of 2 dice X Probability(1,1) 2

(1,2) (2,1) 3

(1,3) (2,2) (3,1) 4

(1,4) (2,3) (3,2) (4,1) 5

6

7

8

9

10

11

12

Outcome of 2 dice X Probability(1,1) 2

(1,2) (2,1) 3

(1,3) (2,2) (3,1) 4

(1,4) (2,3) (3,2) (4,1) 5

(1,5) (2,4) (3,3) (4,2) (5,1) 6

7

8

9

10

11

12

Outcome of 2 dice X Probability(1,1) 2

(1,2) (2,1) 3

(1,3) (2,2) (3,1) 4

(1,4) (2,3) (3,2) (4,1) 5

(1,5) (2,4) (3,3) (4,2) (5,1) 6

(1,6) (2,5) (3,4) (4,3) (5,2) (6,1)

7

8

9

10

11

12

Outcome of 2 dice X Probability(1,1) 2

(1,2) (2,1) 3

(1,3) (2,2) (3,1) 4

(1,4) (2,3) (3,2) (4,1) 5

(1,5) (2,4) (3,3) (4,2) (5,1) 6

(1,6) (2,5) (3,4) (4,3) (5,2) (6,1)

7

(2,6) (3,5) (4,4) (5,3) (6,2) 8

9

10

11

12

Outcome of 2 dice X Probability(1,1) 2

(1,2) (2,1) 3

(1,3) (2,2) (3,1) 4

(1,4) (2,3) (3,2) (4,1) 5

(1,5) (2,4) (3,3) (4,2) (5,1) 6

(1,6) (2,5) (3,4) (4,3) (5,2) (6,1)

7

(2,6) (3,5) (4,4) (5,3) (6,2) 8

(3,6) (4,5) (5,4) (6,3) 9

10

11

12

Outcome of 2 dice X Probability(1,1) 2

(1,2) (2,1) 3

(1,3) (2,2) (3,1) 4

(1,4) (2,3) (3,2) (4,1) 5

(1,5) (2,4) (3,3) (4,2) (5,1) 6

(1,6) (2,5) (3,4) (4,3) (5,2) (6,1)

7

(2,6) (3,5) (4,4) (5,3) (6,2) 8

(3,6) (4,5) (5,4) (6,3) 9

(4,6) (5,5) (6,4) 10

11

12

Outcome of 2 dice X Probability(1,1) 2

(1,2) (2,1) 3

(1,3) (2,2) (3,1) 4

(1,4) (2,3) (3,2) (4,1) 5

(1,5) (2,4) (3,3) (4,2) (5,1) 6

(1,6) (2,5) (3,4) (4,3) (5,2) (6,1)

7

(2,6) (3,5) (4,4) (5,3) (6,2) 8

(3,6) (4,5) (5,4) (6,3) 9

(4,6) (5,5) (6,4) 10

(5,6) (6,5) 11

12

Outcome of 2 dice X Probability(1,1) 2

(1,2) (2,1) 3

(1,3) (2,2) (3,1) 4

(1,4) (2,3) (3,2) (4,1) 5

(1,5) (2,4) (3,3) (4,2) (5,1) 6

(1,6) (2,5) (3,4) (4,3) (5,2) (6,1)

7

(2,6) (3,5) (4,4) (5,3) (6,2) 8

(3,6) (4,5) (5,4) (6,3) 9

(4,6) (5,5) (6,4) 10

(5,6) (6,5) 11

(6,6) 12

1/362/363/364/365/366/365/364/363/362/361/36

12

Probability histogram

13

Roll a die and let X denote the outcome

1. p(X≤2)=

2. p(X<3)=

3. p(2≤X≤4)=

4. p(X>1)=

5. p(X≤6)=

Example

Consider an unbalanced die with the following probabilities:

Value of X Probability1 1/62 1/63 1/64 1/125 1/126 2/6

Roll a die and let X denote the outcome

1. p(X≤2)=p(X=1)+p(X=2)=1/6+1/6=1/3

2. p(X<3)= p(X=1)+p(X=2)=1/6+1/6=1/3

3. p(2≤X≤4)=p(X=2)+p(X=3)+p(X=4)=1/6+1/6+1/12=5/12

4. p(X>1)=1-p(X=1)=1-1/6=5/6

5. p(X≤6)=sum of all probabilities =1

14

Mean and standard deviation of a random variable

15

Mean of a random variable

• The mean of a list of numbers is their average• The mean of a rv X is a weighted average of the possible

values of X.

Example – household size in the U.S

X 1 2 3 4 5 6 7 Probability .251 .321 .171 .154 .067 .022 .014

Mean of X==(1)(.251)+(2)(.321)+(3)(.171) +(4)(.154) +(5)(.067) +(6)(.022) +(7)(.014)==2.587

X

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X x1 x2 x3 …… xk

Probability p1 p2 p3 …… pk

μX = E(X) = x1p1+x2p2+……+xkpk=

Mean = expected value

k

1iii px

Mean of a random variable

17

Example

Roll a die once and let X denote the outcome:

Value of X Probability1 1/62 1/63 1/64 1/65 1/66 1/6

Calculate the mean outcome of the die:

E(X)=μX=

=(1)(1/6)+(2)(1/6)+(3)(1/6)+(4)(1/6)+(5)(1/6)+(6)(1/6)=3.5

18

Standard deviation of a random variable We saw that we can describe a probability distribution by a probability histogram.

For the dice example we had:

In addition of the mean we need a measure of the spread.

- variance of a random variable X - standard deviation of a random variable X

2X

X

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Standard deviation of a random variable Weighted average of the squared deviations of the rv X from its mean μX.

X x1 x2 x3 …… xk

Probability p1 p2 p3 …… pk

= (x1- μX)2P1+(x2- μX)2P2+……+(xk- μX)2Pk=

=Σ (xi- μX)2Pi

2X

i=1

k

20

In the 2 dice example:

μX=2(1/36)+3(2/36)+4(3/36)+5(4/36)+6(5/36)+7(6/36)+

+8(5/36)+9(4/36)+ 10(3/36)+11(2/36)+12(1/36)=7

= Var(X)=(2-7)2(1/36)+(3-7)2(2/36)+ (4-7)2(3/36) +(5-7)2(4/36) +(6-7)2(5/36) +(7-7)2(6/36)+(8-7)2(5/36) +(9-7)2(4/36)+ (10-7)2(3/36)+(11-7)2(2/36)+(12-7)2(1/36)=5.8333

= SD(X)=2.415x

Outcome of 2 dice X Probability(1,1) 2 1/36(1,2) (2,1) 3 2/36(1,3) (2,2) (3,1) 4 3/36(1,4) (2,3) (3,2) (4,1) 5 4/36(1,5) (2,4) (3,3) (4,2) (5,1) 6 5/36(1,6) (2,5) (3,4) (4,3) (5,2) (6,1) 7 6/36(2,6) (3,5) (4,4) (5,3) (6,2) 8 5/36(3,6) (4,5) (5,4) (6,3) 9 4/36(4,6) (5,5) (6,4) 10 3/36(5,6) (6,5) 11 2/36(6,6) 12 1/36

2X

21

QuestionAt a news stand, the daily number X of requests for a certain out-of-town newspaper has the following probability distribution:

Number 0 1 2 3Probability .1 .1 .5 .3

(i) What kind of random variable is X? discrete/continuous(ii) What is the probability of fewer than 2 requests?

p(X<2)=p(X=1)+p(X=0)=0.1+0.1=0.2(iii) What is the expected number of requests?

E(X)= μX=0(.1)+1(.1)+2(.5)+3(.3)=2(iv) What is the standard deviation of number of requests?

SD(x)=(0-2)2(.1) + (1-2)2(.1) + (2-2)2(.5) + (3-2)2(.3)=√.8=.894(v) What is the shape of the distribution of requests?

left skewed / symmetric / right skewed

0 1 2 3

.5

.4

.3

.2

.1

22

The Binomial Distribution

p)B(n,~X

n0,1,2,..., x p)(1px

np(x) xnx

x

p(x)Distribution for p=0.5

23

The Binomial distributionExample

A certain medicine may have some side effects with probability 0.3. The medicine is given to 5 people. Find the probability that exactly 2 of them will suffer for the side effects (the other 3 will not have these side effects!)

1 2 3 4 5

sideeffects

sideeffects

0.3 × 0.3 × 0.7 × 0.7 × 0.7 =

=0.03087

But this is only one possibility of choosing the 2 that have side effects

24

Other possibilities are:

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

Each possibility has a probability of 0.32×0.73=0.03087

10 such possibilities: 10×0.03087=0.3087

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You do not need to count the possibilities–

For a total of n trials (5 trials in the medicine example):If:x is the number of “successes” )e.g., side effects)n-x is the number of “failures” (e.g., no side effects)

Then The number of different ways of getting x successes is:

10)123(12

12345

2)!-(52!

5!

2

5

x)!-(nx!

n!

x

n

In the medicine example:

26

The Binomial distributionA random variable X that counts the number of successes in n independent trials is called a binomial variable.

Notations:n – number of independent trialsSuccess – each trial results in a “success” or “failure”p – probability of success in each trial

X~B(n,p) x=0,1,2,…,n

P(X=x) = px(1-p)n-x

n

x

27

The Binomial distribution

In the medicine example:

n=5

Success =side effects

p=0.3

P(X=2) = 0.32(1-0.3)5-2 = 10×0.32×0.73 = .30875

2

28

ExampleAn ad agency seeks comments on a commercial that aired during TVcoverage of the super bowl. The proportion of all U.S adults watchingwas 0.4. The agency takes a random sample of 6 adults.

1. What is the probability that exactly 4 were watching?

n= “success”= p=

X is the

X~

p(X= )

2. What is the probability that at least 4 were watching?

p(X≥4)=p(X=4) + p(X=5) + p(X=6)

.44.62 + .45.61 + .46.60 = .1382+.0369+.0041=.1792

64

6∙5∙4∙3∙2∙1

(4∙3∙2∙1)(2∙1)

64

65

66

6 watching 0.4

B(6,0.4)

number of people watching the game in a sample of 6 people

4 = .44.62 = (.009216) =.138

29

Using the Binomial table

X~B(6,0.4)

p(X=4)=?

n=6

p=0.4

k=4

p(X=4)=0.1382

p(X≥4)=p(X=4) + p(X=5) + p(X=6)=.1382+.0369+.0041=.1792

n=6

p=0.4

k=4

n=6

p=0.4

k=5

n=6

p=0.4

k=6

30

Using the Binomial table

X~B(5,0.5)P(X=0)=

P(X=2)=

P(X=4)=

X~B(5,0.1)P(X=0)=

P(X=2)=

P(X=4)=

.0313

.3125

.1562

.5905

.0729

.0004

31

How to use the table when p>0.5

X~B(5,0.6)

n=5 p=.6

p(X=2)=?

Instead of p=0.6 we take p=0.4

And look in the table for the probability of failures

p(X=5-2)=p(X=3)=.2304

Note that p(X=2)= .62.43 p(X=3)= .43.62

n=5, p=.4

The table does not

give us p>0.5

n=5, p=.6

52

53

32

How to use the table when p>0.5

X~B(8,0.7)

P(X=5)=

n=8 p=0.7 no p=.7 in the table

Instead, we take:

Then-

Instead of looking for p(X=5) we look for:

P(X=3)= _________

85

n=8, p=.3

p=1-0.7=0.3

p(X=3)

.75.33

.2541

33

Binomial probabilities in Minitab

In the session window:

Cumulative Distribution Function

Binomial with n = 6 and p = 0.400000

x P( X <= x ) 0.00 0.0467 1.00 0.2333 2.00 0.5443 3.00 0.8208 4.00 0.9590 5.00 0.9959 6.00 1.0000

p(X≤4)=0.959

34

Question

X = number of diamonds in 3 random cards that are drawn, without replacement, from a deck of 52 cards.

- Why isn’t X a binomial variable?

The probability of drawing a diamond is not the same in the 3 trials!!!

- How can we redefine X as a binomial variable?

X = number of diamonds in 3 random cards that are drawn, with replacement, from a deck of 52 cards.

35

Binomial mean and SD

X~B(6,0.4)

x P( X = x ) 0.00 0.0467 1.00 0.1866 2.00 0.3110 3.00 0.2765 4.00 0.1382 5.00 0.0369 6.00 0.0041

μx=(0)(.0467) + (1)(.1866) + (2)(.3110) + (3)(.2765) + (4)(.1382) +

+ (5)(.0369) + (6)(.0041) = 2.4

Note that μx= 2.4 = (6)(0.4) = np

36

Binomial mean and SDX~B(6,0.4)

x P( X = x ) 0.00 0.0467 1.00 0.1866 2.00 0.3110 3.00 0.2765 4.00 0.1382 5.00 0.0369 6.00 0.0041

σ x=√ (0-2.4)2(.0467) + (1-2.4)2(.1866) + (2-2.4)2(.3110) + (3-

2.4)2(.2765) + (4-2.4)2(.1382) + (5-2.4)2(.0369) + (6-2.4)2(.0041)

= √ 1.44 = 1.2

Note that σ x= √ 1.44 = √(6)(0.4)(0.6)= √ np(1-p)

37

Binomial mean and SD

X~B(n,p)

p)-np(1 2 x

np E(X)μ x

p)-np(1 x

Mean of X:

Variance of X:

Standard deviation of X:

38

Example – blood type

The probability of having blood type O is 0.5. 4 people are chosen at random.

1. Use the Binomial formula to find the probability that 2 of them have blood type O.

Answer: X= number of people with blood type O in a sample of 4

peoplen=4p=0.5p(X=2)=0.375

39

Example – blood type

2. Use the Binomial table to find the probability that at least 1 of them have blood type O.

Answer:

p(X≥1)=p(x=1)+ p(x=2)+ p(x=3)+ p(x=4) =

= 1- p(X=0)= 1-0.0625

40

Example – blood type

3. Draw the probability histogram of the number of people with blood type O in the sample of 4 people

Answer: x P( X = x ) 0.00 0.0625 1.00 0.2500 2.00 0.3750 3.00 0.2500 4.00 0.0625

0 1 2 3 4

0.1

0.2

0.3

0.4

number of people

prob

abilit

y

41

Example – blood type

4. What is the mean number of people that have blood type O out of the 4 people sampled?

Answer:

μx=E(X)= np = (4)(0.5)=2

42

Example – blood type

5. What is the standard deviation of the number of people that have blood type O out of the 4 people sampled?

Answer:

σ x= √np(1-p) = √(4)(0.5)(0.5)=1

43

Example - internet

According to government data, 25% of employed women have never been married.

(a) If 10 employed women are selected at random, what is the probability that exactly 2 have never been married? n=10 p=.25 p(X=2)=.2816

(b) What is the probability that 2 or fewer have never been married?n=10 p=.25 p(X≤2)=.0563+.1877+.2816

(c) What is the probability that at least 8 have never been married?n=10 p=.25 p(X≥8)=.00004+.0000+.0000

Minitab:..\binomial in class.MPJ

44

The normal approximation to the Binomial variable

N(μ=np,σ2=np(1-p))

B(n,p)

45

0 1 2 3 4 5 6

M&M exampleIn a large bowl of M&M’s, the proportion of blues is 1/6 (or .17).

X- the number of blue M&M’s in a sample of size 6

X~B(6, 1/6)

Draw the probability histogram of X and compute its mean and SD

x p(x)0 1(1/6)0(5/6)6=.331 6(1/6)1(5/6)5=.402 15(1/6)2(5/6)4=.203 20(1/6)3(5/6)3=.054 15(1/6)4(5/6)2=.015 6(1/6)5(5/6)1=.006 1(1/6)6(5/6)0=.00

.5

.4

.3

.2

.1

Shape: Skewed right

Mean: 6(1/6)=1 SD: √6(1/6)(5/6)=.9

46

Suppose we take a sample of size 30 from the M&M bowl

X~B(30, 1/6)

Describe the center, spread, and shape of the distribution of X

x P( X = x ) 0.00 0.0037 1.00 0.0230 2.00 0.0682 . . .

30.00 0.000

Minitab: ..\binomial in class.MPJ

47

3029282726252423222120191817161514131211109876543210

0.2

0.1

0.0

X

p(x

)

Shape: Smoother, more bell shaped

Mean: 30(1/6)=5 SD: √30(1/6)(5/6)=2

X~B(30, 1/6)

48

Suppose we take a sample of size 90 from the M&M bowl

X~B(90, 1/6)

Describe the center, spread, and shape of the distribution of X

x P( X = x ) 0.00 0.0000 1.00 0.0000 2.00 0.0000 3.00 0.0001 4.00 0.0002 . . 90.00 0.000

49

9089888786858483828180797877767574737271706968676665646362616059585756555453525150494847464544434241403938373635343332313029282726252423222120191817161514131211109876543210

0.10

0.05

0.00

X

p(x

)

Shape: Even smoother, more bell shaped – very close to a normal curve

Mean: 90(1/6)=15 SD: √90(1/6)(5/6)=3.5

X~B(90, 1/6)

50

The binomial variable X~B(90, 1/6) behaves approximately like a normal variable with mean 15 and SD 3.5

51

The Normal approximation to the Binomial distribution

As sample size n gets large, the distribution of the binomial random variable X is well approximated by the normal distribution, with the same mean and SD of the binomial variable:

If X~B(n,p)

as n increases X~>N(μ=np, σ=√np(1-p))

52

The Normal approximation to the Binomial distribution

How large should n be?

This depends on the value of p.

If p is close to 0.5, then the normal approximation applies for small values of n

If p is far from 0.5, larger values of n are needed.

The following rule of thumb helps us decide when to use the normal approximation:

np≥15 and n(1-p) ≥15

53

The Normal approximation to the Binomial distribution

How large should n be?

This depends on the value of p.

If p is close to 0.5, then the normal approximation applies for small values of n

If p is far from 0.5, larger values of n are needed.

The following rule of thumb helps us decide when to use the normal approximation:

np≥15 and n(1-p) ≥15

54

Example

Check if the rule of thumb is satisfied for:

1. B(n=6, p=1/6)

no, since np=6(1/6)=1<15

2. B(n=90, p=1/6)

yes, since np=90(1/6)=15 and n(1-p)=90(5/6)=75>15

55

Example of normal approximation to binomial

You operate a restaurant. You read that sample survey by the National Restaurant association shows that 40% of adults are committed to eating nutritious food when eating away from home. To help plan your menu, you decide to conduct a sample survey in your own area. You will use random digit dialing to contact an SRS of 200 households by telephone. If the national results hold in your area, it is reasonable to use B(200,0.4) distribution to describe the count X of respondents who seek nutritious food when eating out.

(a) What is the mean number of nutrition-conscious people in your

sample if p=.4 is true? Mean of B(200,.4)=200(.4)=80 SD=√200(.4)(.6)=√48=6.93

(b) Use the normal approximation to compute the probability that X lies between 75 and 85

56

Is the normal approximation appropriate?

np=200(.4)=80>15 n(1-p)=200(.6)=120>18

X~>N(80,6.932)

5284.72.72. Zp

93.6

8085

93.6

80

93.6

8075 Xpp(75≤X≤85) =

57

The normal approximation applies also for proportions

Next we will show that if X~B(n,p),

As n increases –

is approximately N( )n

p)-p(1σ p,μ

n

Xp̂

58

The normal approximation applies also for proportion

Since np=500(.45)=225 and n(1-p)=500(.55)=275 we can use the normal approximation:

μX=np=225

σX=√ np(1-p)=√500(.45)(.55)=√123.75 = 11.12X~>N(225, 11.122)

A poll that surveyed 500 people found that 45% of them support military action in Iraq.X – the number of people that support military action in Iraq in a sample of 500 people

X~B(500,0.45)

Is the normal approximation appropriate in this case?

59

Lets define a new variable:

- the proportion of people that support military action in Iraq

Instead of looking at the number of people who support that attack, we can look at the proportion of people who support the attack.

Denote this proportion by

How does behave?

If X is Normal, any linear transformation of X is also normal.

Since is a linear transformation of X -

is approximately normal (note that X is approximately normal)

Calculating the mean and SD of requires some definitions

p̂

Xn

Xn

p̂

n

Xp̂

p̂

p̂

60

Mean and SD of a linear transformation of a random variable

If X is a random variable

and a and b are fixed numbers:

μa+bX=E(a+bX)=a+bμX

σ2a+bX=Var(a+bX)=b2σ2

X

61

Mean and variance of

n

p)-p(1p)-np(1

n

1Var(X)

n

1

n

XVar)p̂Var(

22

pnpn

1E(X)

n

1)

n

XE()p̂E(

n

Xp̂

62

p-np(1σ np,μNely approximat is X

n

p)-p(1σ p,μNely approximat is

n

Xp̂

If X~B(n,p)

as n increases (rule of thumb: np≥15 and n(1-p) ≥15)

Summary - Normal approximation to the Binomial distribution

63

Back to the example

n

p)-p(1σ p,μNely approximat is p̂

)0.022 N(.4,ely approximat is p̂ 2

μ = p = 0.4

σ = 022.000495.500

)55.(.45)(

n

p)p(1

64

Example of normal approximation to binomial

You operate a restaurant. You read that sample survey by the National Restaurant association shows that 40% of adults are committed to eating nutritious food when eating away from home. To help plan your menu, you decide to conduct a sample survey in your own area. You will use random digit dialing to contact an SRS of 200 households by telephone. If the national results hold in your area, it is reasonable to use B(200,0.4) distribution to describe the count X of respondents who seek nutritious food when eating out.

(a) What is the mean number of nutrition-conscious people in your sample if p=.4 is true?

(b) Define by p-the proportion of people in the sample that seek nutritious food when eating out. p=X/n.Use the normal approximation to compute the probability that the p is larger than 0.7.

65

X~>N(80,6.932)

0032.9968.1)73.2(1)73.2(

11.

4.7.

11.

4.ˆ7.ˆ

Zp

pppp

11.,4.~ˆ

200

)6(.4.,4.~ˆ

)1(

,~ˆ

Nn

Xp

Nn

Xp

n

pppN

n

Xp